How do you solve $x^{2} - 2 x - 5 = 0$ $x^2-2x-5=0$ by completing the square?

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How do you solve $x^{2} - 2 x - 5 = 0$ $x^2-2x-5=0$ by completing the square?

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Answer 1 · 2015-06-30T04:22:56

Solve $x^{2} - 2 x - 5$ $x^2 - 2x - 5$

Explanation:

$x^{2} - 2 x = 5$ $x^2 - 2x = 5$
$x^{2} - 2 x + 1 = 5 + 1 = 6$ $x^2 - 2x + 1 = 5 + 1 = 6$
${(x + 1)}^{2} = 6$ $(x +1)^2 = 6$
$(x + 1) = \pm \sqrt{6}$ $(x + 1) = +- sqrt6$

$x = - 1 \pm \sqrt{6}$ $x = -1 +- sqrt6$

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How do you solve $x^{2} - 2 x - 5 = 0$ $x^2-2x-5=0$ by completing the square?

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Explanation:

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How do you solve $x^{2} - 2 x - 5 = 0$ $x^2-2x-5=0$ by completing the square?